Problem: $ D = \left[\begin{array}{rr}5 & 1 \\ -1 & 3\end{array}\right]$ $ A = \left[\begin{array}{rr}2 & -1 \\ 1 & 4\end{array}\right]$ What is $ D A$ ?
Answer: Because $ D$ has dimensions $(2\times2)$ and $ A$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ D A = \left[\begin{array}{rr}{5} & {1} \\ {-1} & {3}\end{array}\right] \left[\begin{array}{rr}{2} & \color{#DF0030}{-1} \\ {1} & \color{#DF0030}{4}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ D$ , with the corresponding elements in column $j$ of the second matrix, $ A$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ D$ with the first element in ${\text{column }1}$ of $ A$ , then multiply the second element in ${\text{row }1}$ of $ D$ with the second element in ${\text{column }1}$ of $ A$ , and so on. Add the products together. $ \left[\begin{array}{rr}{5}\cdot{2}+{1}\cdot{1} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ D$ with the corresponding elements in ${\text{column }1}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{5}\cdot{2}+{1}\cdot{1} & ? \\ {-1}\cdot{2}+{3}\cdot{1} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ D$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{5}\cdot{2}+{1}\cdot{1} & {5}\cdot\color{#DF0030}{-1}+{1}\cdot\color{#DF0030}{4} \\ {-1}\cdot{2}+{3}\cdot{1} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{5}\cdot{2}+{1}\cdot{1} & {5}\cdot\color{#DF0030}{-1}+{1}\cdot\color{#DF0030}{4} \\ {-1}\cdot{2}+{3}\cdot{1} & {-1}\cdot\color{#DF0030}{-1}+{3}\cdot\color{#DF0030}{4}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}11 & -1 \\ 1 & 13\end{array}\right] $